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Expert-verified Found in: Page 433 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # In problems 1-6, determine the convergence set of the given power series.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{\right)}\mathbf{!}}{\mathbf{n}\mathbf{!}}{\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{\right)}}^{{\mathbf{n}}}$

The set is, $x\in \left(-3,-1\right)$

See the step by step solution

## Step 1:To Find the Radius of convergence

Use the ratio test to determine the radius of convergence (with ):${a}_{n}=\frac{\left(n+2\right)!}{n!}$

$\begin{array}{c}\underset{n\to \infty }{\mathrm{lim}}\left|\frac{{a}_{n}}{{a}_{n+1}}\right|=\underset{n\to \infty }{\mathrm{lim}}\left|\frac{\frac{\left(n-2\right)!}{n!}}{\frac{\left(n+3\right)!}{\left(n+1\right)!}}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{\left(n+2\right)!}{\left(n+3\right)!}\cdot \frac{\left(n+1\right)!}{n!}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{\left(n+2\right)!}{\left(n+3\right)\left(n+2\right)!}×\frac{\left(n+1\right)n!}{n!}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{\left(n+1\right)}{n+3}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{\left(1+\left(1/n\right)\right)}{1+\left(3/n\right)}\right|\\ =1\end{array}$

The radius of convergence is 1, therefore convergent set for the given power series is.$|x+2|<1$

## Step 2: Find the set of convergence

To completely identify the convergence set, we have to check whether the boundary points -1 and -3 are included in the set or not.

Checking at$x=-1$ , by substituting the value ofx by -1 .

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{\left(n+2\right)!}{n!}{\left(x+2\right)}^{n}=\sum _{n=0}^{\infty }\frac{\left(n+2\right)!}{n!}{\left(-1+2\right)}^{n}\\ =\sum _{n=0}^{\infty }\frac{\left(n+2\right)!}{n!}{\left(1\right)}^{n}\\ =\sum _{n=0}^{\infty }\frac{\left(n+2\right)\left(n+1\right)n!}{n!}\\ =\sum _{n=0}^{\infty }\left(n+2\right)\left(n+1\right)\\ =\infty \end{array}$

The above series is divergent, thus the point -1 is excluded from the convergent set

Similarly, checking at $x=-3$ by substituting the value of x by -3.

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{\left(n+2\right)!}{n!}{\left(x+2\right)}^{n}=\sum _{n=0}^{\infty }\frac{\left(n+2\right)!}{n!}{\left(-3+2\right)}^{n}\\ =\sum _{n=0}^{\infty }\frac{\left(n+2\right)!}{n!}\\ =\sum _{n=0}^{\infty }\frac{\left(n+2\right)\left(n+1\right)n!}{n!}\\ =\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\left(n+2\right)\left(n+1\right)\end{array}$

The above series is divergent, thus the point -3 is not included in the convergent set. The convergent set for the given power series is.$x\in \left(-3,-1\right)$ ### Want to see more solutions like these? 